When is $F(x+y) = F(x,y)$ for field $F$?

Question

If $F$ is a field and $x,y$ are in an algebraic extension of $F$, I'm curious as to what we can say about $[F(x+y):F]$. I can easily prove the following:

  $[F(x+y):F] \mid [F(x,y):F]$

  $[F(x+y):F] \ge \max([F(x,y):F(x)],[F(x,y):F(y)])$

  $[F(x+y):F]^2 \ge \dfrac{[F(x,y):F]}{\gcd([F(x):F],[F(y):F])}$

But can more be said? And are there any simple conditions that ensure that $F(x+y) = F(x,y)$? For example, if $\gcd([F(x):F],[F(y):F]) = 1$, is it always the case? I doubt, but don't know how to construct a counter-example.

Answer 1

This is essentially the comment by KCd, answering my conjecture affirmatively if $\text{char}(F)=0$.

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As stated in this MathOverflow post, citing Isaac's paper, we have:

If $\text{char}(F)=0$ and $[F(x,y):F] = [F(x):F]·[F(y):F]$ then $F(x+y) = F(x,y)$.

In particular this implies but is strictly stronger than:

If $\text{char}(F)=0$ and $\gcd([F(x):F],[F(y):F]) = 1$ then $F(x+y) = F(x,y)$.

Another post on the same MO thread gives a counter-example to the conjecture if $\text{char}(F) \ne 0$: $ \def\ff{\mathbb{F}} $

Let $s,t$ be indeterminates and $K = \ff_p(s,t)$ for prime $p > 2$. Let $a,b$ be algebraic over $K$ such that $a\,^{p-1}-s = 0$ and $b\,^p-s·b-t = 0$. Then $[K(a):K] = p-1$ and $[K(b):K] = p$ but $[K(a+b):K] = p$ because $(a+b)^p-s·(a+b)-t = 0$.

Answer 2

If $F$ is separable then you have a good shot at it. For convenience I will replace $x$ and $y$ by $a$ and $b$ respectively. In particular,

let $f_a,f_b$ be the minimal polynomials for $a$ and $b$ and $\{a=a_1,a_2,\ldots a_m\}$ and $\{b = b_1,b_2,\ldots b_n\}$ be the distinct roots of these in $F(a,b)$.

if $$1 \neq \frac{a_i-a}{b - b_j} \quad \forall i\geq1\quad \forall j>1$$

then from the proof of the Primitive Element Theorem we have:

$$F(a,b) = F(a+b)$$